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# I am posting this as a note because I do not know the exact answer for this...Do post the solutions if u can crack this one!!

A set of positive integers satisfies the property that when 10^{20} , 15^{10} and 24^{15} are divided by any number in this set, at least one of the remainders is zero. What is the total number of elements in this set? a)1256 b)1266 c)1024 d)none of these

Note by Manu Mehta
2 years, 8 months ago

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Let A denote numbers divisible by $$10^{20}=2^{20}5^{20}$$, B denote numbers divisible by $$15^{10}=3^{10}5^{10}$$ and C denote numbers divisible by $$24^{15}=2^{45}3^{15}$$.

What you want is $$a+b+c+d+e+f+g$$.

g=Numbers divisible by all 3. So g has one number {1} so g=$$\color{blue}{1}$$

d+g=Numbers in A and C= divisible by $$2^{20}$$. Hence d+g=21 or d=$$\color{blue}{20}$$

f+g=Numbers in B and C= divisible by $$3^{10}$$. Hence f+g=11 or f=$$\color{blue}{10}$$

e+g=Numbers in A and B= divisible by $$5^{10}$$. Hence e+g=11 or e=$$\color{blue}{10}$$

a+d+e+g=Numbers in A=divisible by $$10^{20}$$. a+d+e+g=441 or a=$$\color{blue}{410}$$

b+f+e+g=Numbers in B=divisible by $$15^{10}$$. b+f+e+g=121 or b=$$\color{blue}{100}$$

c+d+f+g=Numbers in C=divisible by $$24^{15}$$. c+d+f+g=736 or c=$$\color{blue}{705}$$

Adding, we get a+b+c+d+e+f+g=$$\color{red}{\boxed{1256}}$$ · 2 years, 5 months ago